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Question #44050

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The Question

A $0.1\text{ M}$ aqueous solution of a weak monoprotic acid, \ce{HA}, exhibits an osmotic pressure of $2.75\text{ atm}$ at $300\text{ K}$ and $3.60\text{ atm}$ at $360\text{ K}$ . Assuming the volume of the solution is independent of temperature, calculate the standard enthalpy of dissociation ( $\Delta H^\circ$ ) of the acid in $\text{kJ/mol}$ .

[Given: $R = \frac{1}{12}\text{ L atm K}^{-1}\text{ mol}^{-1}$ for osmotic pressure, $R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}$ for thermodynamic calculations. $\ln(2) = 0.70$ , $\ln(3) = 1.10$ , $\ln(4.5) = 1.50$ ]

Options

A.
$22.41\text{ kJ/mol}$
B.
$20.92\text{ kJ/mol}$
C.
$10.46\text{ kJ/mol}$
D.
$1.30\text{ kJ/mol}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Gibbs-Helmholtz Equation|Degree of Dissociation (α) approximation|Molecular Mass Abnormalities

Why This Question Is Tricky

Common Mistakes

This problem tests the integration of colligative properties with chemical thermodynamics.

The primary trap is the 'Degree of Dissociation Approximation'.

Students often assume $1-\alpha \approx 1$ for weak acids, which is only valid for $\alpha < 0.05$ .

Here, $\alpha = 0.1$ and $0.2$ , making the approximation invalid and leading to the incorrect $20.92\text{ kJ/mol}$ .

Secondary traps involve fundamental misunderstandings of the equilibrium constant's relationship with the van't Hoff factor, leading students to incorrectly substitute the ratio of $i$ or $\alpha$ directly into the Gibbs-Helmholtz equation instead of calculating the true $K_{eq}$ .

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Thermodynamics - Class 11 Part 1
Concept	Gibbs-Helmholtz Equation
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free